I am currently dealing with Linear Kinetic Equations of the type $$ \partial_t f + v \cdot \nabla_x f = Qf $$ $Qf $ should be a bounded collision operator but is not important for this question. We consider $t>0 \, (\, in \, \mathbb{R}), x \in \mathbb{T}^d, v \in \mathbb{R}^d,$ where $\mathbb{T}^d$ is the d-dimensional Torus represented by $[0,1]^d$. The function $f=f(x,v,t)$ should be $C^1$ w.r.t to t and $f(x,v) \in \mathcal{H} := L^2(\mathbb{T}^d \times \mathbb{R}^d,dxdvM_T(v)^{-1})$ (a weighted $L^2$-Hilbertspace, $M_T(v)= (2 \pi T)^{-d/2}e^{-|v|^2/(2T)}$ is the Maxwellian and $T > 0$ a constant Temperature) with scalar product $$ \left<f,g \right>_{\mathcal{H}} := \int_{\mathbb{T}^d \times \mathbb{R}^d} \frac{f(x,v)g(x,v)}{M_T(v)} \, d(x,v). $$
Now, I often read that the Transport Operator $T:= v \cdot \nabla_x $ is skew-symmetric on $\mathcal{H}$ which is clear for me through Integration by parts if I propose periodic boundary conditions on $f$ w.r.t to $x$. But is it still possible to deduce this without assuming periodic boundary conditions?
For simplicity let us consider the one-dimensional case, $d=1$, so the Transport operator becomes $T:= v \cdot \partial_x$ and to show skew-symmetry it is sufficient to show $\left<Tf,f \right>_{\mathcal{H}} = 0$ for all $f \in dom(T) \subseteq \mathcal{H}.$ So by Fubini we can say $$ \left<Tf,f \right>_{\mathcal{H}} = \int_{\mathbb{T}^1 \times \mathbb{R}} \frac{Tf(x,v)f(x,v)}{M_T(v)} \, d(x,v) $$ $$= \int_{\mathbb{R}} \int_{\mathbb{T}^1} \frac{v \cdot \partial_xf(x,v)f(x,v)}{M_T(v)} \, dxdv $$ $$= \frac{1}{2}\int_{\mathbb{R}} \frac{v}{M_T(v)}\left (\int_{\mathbb{T}^1} \partial_xf^2(x,v) \, dx \right)dv$$ $$ = \frac{1}{2}\int_{\mathbb{R}} \frac{v}{M_T(v)}\left (f^2(1,v)-f^2(0,v) \right)dv.$$ My problem is now how do I get on here without using periodic boundary conditions? I know $$ \int_{\mathbb{R}} \frac{v}{M_T(v)} dv = 0$$ by symmetry of the integrand but how can I use this here?
Would be very grateful for any help!